# Pushforwards in local coordinates

1. Feb 28, 2016

### Math Amateur

I am reading John M. Lee's book: Introduction to Smooth Manifolds ...

I am focused on Chapter 3: Tangent Vectors ...

I need some help in fully understanding Lee's conversation on computations with tangent vectors and pushforwards ... in particular I need help with an aspect of Lee's exposition of pushforwards in coordinates ... ...

The relevant conversation in Lee is as follows:

In the above text we read:

" ... ... Thus

$F_* \frac{ \partial }{ \partial x^i } |_p = \frac{ \partial F^j }{ \partial x^i } (p) \frac{ \partial }{ \partial y^j } |_{ F(p) }$ ... ... ... ... 3.6

In other words, the matrix of $F_*$ in terms of the standard coordinate basis is

$\begin{pmatrix} \frac{ \partial F^1 }{ \partial x^1 } (p) & ... & ... & \frac{ \partial F^1 }{ \partial x^n } (p) \\ ... & ... & ... & ... \\ ... & ... & ... & ... \\ \frac{ \partial F^m }{ \partial x^1 } (p) & ... & ... & \frac{ \partial F^m }{ \partial x^n } (p) \end{pmatrix}$

... ... ... "

My question is as follows:

How ... exactly ... do we get from equation 3.6 above to the fact that the matrix of [MATH]F_*[/MATH] in terms of the standard coordinate basis is

$\begin{pmatrix} \frac{ \partial F^1 }{ \partial x^1 } (p) & ... & ... & \frac{ \partial F^1 }{ \partial x^n } (p) \\ ... & ... & ... & ... \\ ... & ... & ... & ... \\ \frac{ \partial F^m }{ \partial x^1 } (p) & ... & ... & \frac{ \partial F^m }{ \partial x^n } (p) \end{pmatrix}$

... ... ... ?

It looks as if Lee derives $(F_*)_{ij} = \frac{ \partial F^j }{ \partial x^i } (p)$ from 3.6 ... but how exactly is this justified ... that is, what are the mechanics of this ... I cannot see it ... can someone please help ... ..

Peter

*** EDIT ***

It has occurred to me that it would be helpful for readers of the post to have access to Lee's definition of pushforwards, and his early remarks on the properties of pushforwards ... ... so I am providing these as follows:

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• ###### Lee - Definition and Properties of the Pushforward ... ....png
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2. Feb 29, 2016

### Math Amateur

I really should not correspond with and talk with my self ... ... ... ... but anyway ... have been reflecting on my post above ...

Equation 3.6, I have realised uses what Lee calls the Einstein Summation Convention and so reads

$F_* \frac{ \partial }{ \partial x^i } |_p = \sum_j \frac{ \partial F^j }{ \partial x^i } (p) \frac{ \partial }{ \partial y^j } |_{ F(p) }$ ... ... ... ... 3.6

and so for i = 1 we have

$F_* \frac{ \partial }{ \partial x^i } |_p = \frac{ \partial F^1 }{ \partial x^1 } (p) \frac{ \partial }{ \partial y^1 } |_{ F(p) } + \frac{ \partial F^2 }{ \partial x^1 } (p) \frac{ \partial }{ \partial y^2 } |_{ F(p) } + \frac{ \partial F^3 }{ \partial x^1 } (p) \frac{ \partial }{ \partial y^3 } |_{ F(p) } + \ ... \ ... \ + \frac{ \partial F^m }{ \partial x^i } (p) \frac{ \partial }{ \partial y^m } |_{ F(p) }$

$= ( \frac{ \partial }{ \partial y^1 } |_{ F(p) }, \frac{ \partial }{ \partial y^2 } |_{ F(p) } , \frac{ \partial }{ \partial y^3 } |_{ F(p) }, \ ... \ ... \ , \frac{ \partial }{ \partial y^m } |_{ F(p) } )$

$\times$

$\begin{pmatrix} \frac{ \partial F^1 }{ \partial x^1 } (p) \\ \frac{ \partial F^2 }{ \partial x^1 } (p) \\ \frac{ \partial F^3 }{ \partial x^1 } (p) \\ ... \\ ... \\ \frac{ \partial F^m }{ \partial x^1 } (p) \end{pmatrix}$

The column vector above is the first column vector of the required Jacobian ... ... taking $i = 2$ and proceeding in the same way gives column $2$ and so on ...

Is that correct?

Peter

3. Feb 29, 2016

### Brian T

Yes that look correct. To see where it comes from:

Think about what I said in my other reply to your previous post; you can think of any vector $v$ as the tangent vector of a curve $\alpha(t)$ passing through $p \in M$ at $t=0$, which when applied to a function gives the time derivative of that function.
$$v |_{p} = \frac{dx^j}{dt} |_{t=0} \frac{\partial}{\partial x^j}|_p$$
Now, consider a smooth map, $F: M \rightarrow N$. Keeping with this curve-vector idea, we define the pushforward as the tangent vector to the corresponding curve on $N$. Taking $\alpha: \mathbb{R} \rightarrow M$ to be a curve on M, $F \circ \alpha: \mathbb{R} \rightarrow N$ is the corresponding curve on N.

By definition, the pushforward of $v$ is just the tangent vector to the new curve, which we construct as usual (take the derivatives of the components of the curve, which we'll call $F^j$, and multiply by the basis vectors on $N$:
$$F_*(v)|_{F(p)} = \frac{ d(F^j \circ \alpha) }{dt} |_{t=0} \frac{\partial}{\partial y^j} |_{F(p)}$$
Using the chain rule, on the first part of the right hand side, (and dropping the evaluation symbols)
$$F_*(v) = \frac{\partial(F^j \circ \alpha) }{\partial x^k}\frac{dx^k}{dt} \frac{\partial}{\partial y^j}$$

Comparing this expression to our original expression for $v$, we see that we get the additional Jacobian term and a new basis. If you want to reduce the relation of the pushforward of $v$ to just the pushforward on the basis vectors, take $v = \frac{\partial}{\partial x^j}$ (for a given value of j). The pullback is simply our previous derived expression but take away the components (set the components =1)
$$F_*\frac{\partial}{\partial x^j} = \frac{\partial (F^i \circ \alpha) }{\partial x^j} \frac{\partial}{\partial y^i}$$

4. Feb 29, 2016

### Math Amateur

Thanks Brian ... still reflecting on that ...

What text do you use for this theory ... ?

Peter

5. Feb 29, 2016

### Brian T

Hey Peter,
As with most topics, I usually synthesize from various sources. I prefer this since you get exposed to various different explanations of a given concept (and become more familiar with a variety of notation). I would say that the main texts I'm reading through right now are Riemannian Geometry by Manfredo do Carmo and Differentiable Manifolds: A Theoretical Physics Approach by Torres del Castillo. Also, if you like seeing how geometry is applied to relativity, I'd recommend Wald's General Relativity.

6. Feb 29, 2016

### WWGD

I am not sure I understand your notation, but the pullback and the pushforward go in opposite directions.

7. Feb 29, 2016

### Brian T

Ah thanks. I typed pullback but meant pushforward

Specifically, we derived an expression for the pushforward of any vector (in coordinate basis), so the pushforward of just the basis vectors is a simple instance of the more general expression